Given, equation of circle
$$
x^2+y^2-6 x+4 y-12=0
$$
$$
(x-3)^2+(y+2)^2=(5)^2
$$
parametric form of circle is
$$
\begin{aligned}
& x=3+5 \cos \theta, y=-2+5 \sin \theta \\
& \text { When } \theta=30^{\circ} \\
& \quad A\left(3+5 \cos 30^{\circ},-2+5 \sin 30^{\circ}\right) \\
& \quad A\left(3+\frac{5 \sqrt{3}}{2},-2+\frac{5}{2}\right)=\left(\frac{6+5 \sqrt{3}}{2}, \frac{1}{2}\right)
\end{aligned}
$$
$$
\begin{aligned}
& \text { When } \theta=90^{\circ} \\
& \qquad \begin{array}{l}
B\left(3+5 \cos 90^{\circ},-2+5 \sin 90^{\circ}\right) \\
B(3,3)
\end{array}
\end{aligned}
$$
$\therefore$ Equation of chord $A B$ where
$$
\begin{aligned}
& A=\left(\frac{6+5 \sqrt{3}}{2}, \frac{1}{2}\right) \text { and } B=(3,3) \\
& y-3=\frac{\frac{1}{2}-3}{\frac{6+5 \sqrt{3}}{2}-3}(x-3) \\
\Rightarrow & y-3=\frac{-1}{\sqrt{3}}(x-3) \Rightarrow \sqrt{3} y-3 \sqrt{3}=-x+3 \\
\Rightarrow & x+\sqrt{3} y-3 \sqrt{3}-3=0 \\
\Rightarrow & x+\sqrt{3} y-3(1+\sqrt{3})=0
\end{aligned}
$$